Is ${749628}$ divisible by $9$ ?
Explanation: A number is divisible by $9$ if the sum of its digits is divisible by $9$ . [ Why? First, we can break the number up by place value: $ \begin{eqnarray} {749628}= &&{7}\cdot100000+ \\&&{4}\cdot10000+ \\&&{9}\cdot1000+ \\&&{6}\cdot100+ \\&&{2}\cdot10+ \\&&{8}\cdot1 \end{eqnarray} $ Next, we can rewrite each of the place values as $1$ plus a bunch of $9$ s: $ \begin{eqnarray} {749628}= &&{7}(99999+1)+ \\&&{4}(9999+1)+ \\&&{9}(999+1)+ \\&&{6}(99+1)+ \\&&{2}(9+1)+ \\&&{8} \end{eqnarray} $ Now if we distribute and rearrange, we get this: $ \begin{eqnarray} {749628}= &&\gray{7\cdot99999}+ \\&&\gray{4\cdot9999}+ \\&&\gray{9\cdot999}+ \\&&\gray{6\cdot99}+ \\&&\gray{2\cdot9}+ \\&& {7}+{4}+{9}+{6}+{2}+{8} \end{eqnarray} $ Any number consisting only of $9$ s is a multiple of $9$ , so the first five terms must all be multiples of $9$ That means that to figure out whether the original number is divisible by $9 $ , all we need to do is add up the digits and see if the sum is divisible by $9$ . In other words, ${749628}$ is divisible by $9$ if ${ 7}+{4}+{9}+{6}+{2}+{8}$ is divisible by $9$ Add the digits of ${749628}$ $ {7}+{4}+{9}+{6}+{2}+{8} = {36} $ If ${36}$ is divisible by $9$ , then ${749628}$ must also be divisible by $9$ ${36}$ is divisible by $9$, therefore ${749628}$ must also be divisible by $9$.